More on Matrices Representing Complex Numbers This question is in a sense a follow-up and extension of this one, which essentially asks for representations of complex numbers $a + bi \in \Bbb C$ as $2 \times 2$ real matrices $A$ such that
$AJ = JA, \tag 1$
with
$J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}; \tag 2$
we observe that
$J^2 = - I. \tag 3$
The answer I gave there showed that all such matrices are given by
$A \in aI + bJ \in M_{2 \times 2}(\Bbb R); \tag 4$
here we clearly have
$a, b \in \Bbb R, \tag 5$
and the correspondence 'twixt complex numbers and such matrices is given by
$\Bbb C \ni a + bi \longleftrightarrow aI + bJ \in M_{2 \times 2}(\Bbb R). \tag 6$
Seeking a generalization of these results, I here request that 
1.)  All real $2 \times 2$ matrices satisfying (3) be found;
2.)  For any such $J$ as in request (1.), all real matrices $A$ such that (1) binds be found;
3.)  For any such $J$, $A$ as in request (2.), 
$A = aI + bJ, \tag 7$
for some $a, b \in \Bbb R$ be shown;
4.)  The mapping
$a + bi \longleftrightarrow aI + bJ \tag 8$
is an isomorphism 'twixt the complex numbers $\Bbb C$ and the set of $2 \times 2$ real matrices of the form $aI + bJ$.
 A: Let us define $Z=\begin{bmatrix}0& -1\\ 1&0\end{bmatrix}$ for simplicity.
[1] Directly, any matrix of the following form for any real $a,k$:
$$J=\begin{bmatrix}a& -(a^2+1)/k\\ k&-a\end{bmatrix}$$
For completeness, consider a matrix
$$J=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
such that $J^2=-I$. We can directly solve this: $a^2+bc =-1;\ d^2+bc = -1;\ b(a+d)=0;\ c(a+d)=0$. If $a+d$ is non-zero, then $b=c=0$ by the last two equations, and no solution exists for $a^2=-1$. Otherwise, we have $d=-a$, and can parametrize the solutions as done above.
More elegantly, for any invertible $X$,
$$J=\pm X^{-1}Z X$$
fits in (3); for positive $k$, we can just plug in
$$X=\begin{bmatrix}a/\sqrt{k}& 1/\sqrt{k}\\ \sqrt{k}&0\end{bmatrix}$$
to recover the above characterization, so this covers all possible $J$
[2] (Using conjugacy classes, this is essentially the same as the case with $J=Z$ itself.)
More precisely, if $A$ commutes with $J$, let $B=XAX^{-1}$ (i.e. $A=X^{-1}BX$), then
$$AX^{-1}Z X = X^{-1} B Z X$$
$$X^{-1}Z XA = X^{-1} Z B X$$
And these two are only equal when $B$ commutes with $Z$. Therefore, relying on your previous results, the commuting matrices are
$$A=X^{-1}(aI+bZ)X$$
[3] Directly follows from the above.
[4] This should be simple to check, given the above properties. $I$ and $J$ are always linearly independent because the trace of one of them is zero and the other is not.
A: Not exactly an answer, I have added an additional condition. Thanks to @ancientmathematician for catching my mistake.
Suppose we have a continuous injective ring homomorphism $\phi: \mathbb{C} \to \mathbb{R}^{2 \times 2}$.
Let $J= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.
Then I claim that $\phi(a+ib) = aI + b W J W^{-1}$ for some invertible $W$.
We must have $\phi(1) = I$, and so $\phi(q) = qI$ for $q \in \mathbb{Q}$.
Let $B=\phi(i)$. Similarly we have $\phi(qi) = qB$. 
We need to determine the allowable values of $B$.
We have $\phi(i^2) = \phi(-1) = \phi(i)^2$ so $B^2+I = 0$ and $B$ is real and hence
has distinct eigenvalues $\pm i$. Hence for some real $u,v \in \mathbb{R}^2$ we have
$B(u+iv) = i(u+iv) = -v+iu$ and so $Bu = -v, Bv = u$. (It is straightforward to show that $u,v$ are linearly independent.) If we let $W = \begin{bmatrix} u & v\end{bmatrix}$
then $B W = W J$, or
$B = W J W^{-1}$.
Since $\mathbb{Q}[i]$ is dense in $\mathbb{C}$, it follows by continuity that $\phi(a+ib) = aI + b W J W^{-1}$ for any $a,b \in \mathbb{R}$.
It is straightforward to verify that $\phi(a+ib) = aI + b W J W^{-1}$ for any $a,b \in \mathbb{R}$ defines a continuous injective homomorphism.
A: I'll show $(3)$ directly. Let's think of $A$ and $J$ as complex matrices (that is, as matrices in $M_2(\mathbb{C})$). Since $J^2 = -1$, it has $\pm i$ as eigenvalues and is diagonalizable (as a complex matrix). Denote by $v_1,v_2 \in \mathbb{C}^2$ the corresponding eigenvectors so that $Jv_1 = iv_1, Jv_2 = -iv_2$. Since $AJ = JA$, it is readily seen that $v_1,v_2$ are also eigenvectors of $A$. Now, since $A$ is real, we have two possible cases:


*

*$A$ has distinct real eigenvalues. But then $A$ is diagonalizable over $\mathbb{R}$ and any real eigenvector of $A$ will also be an eigenvector of $J$ which is impossible.

*$A$ has eigenvalues $a \pm ib$ with $a,b \in \mathbb{R}$. Then $Av_1 = (a + ib)v_1$ and $Av_2 = (a - ib)v_2$. This immediately implies that $A = aI + bJ$ (apply both sides to $v_1,v_2$).

